Exponentiation by squaring

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Exponentiating by squaring is an algorithm used for the fast computation of large integer powers of a number. It is also known as the square-and-multiply algorithm or binary exponentiation. In additive groups the appropriate name is double-and-add algorithm. It implicitly uses the binary expansion of the exponent. It is of quite general use, for example in modular arithmetic.

Contents 1 Squaring algorithm 2 Further applications 3 Example implementations 3.1 Computation by powers of 2 3.1.1 Runtime example: Compute 310 3.2 Runtime example: Compute 310 = 3.3 Generalization with example 3.3.1 Generalization 3.3.2 Text application 3.4 Calculation of products of powers 3.4.1 Example 3.4.2 Using transformation 3.4.3 Examples 3.4.4 Implementation 4 Alternatives

[edit] Squaring algorithm

The following recursive algorithm computes xⁿ for a positive integer n:

Compared to the ordinary method of multiplying x with itself n − 1 times, this algorithm uses only O(log n) multiplications and therefore speeds up the computation of xⁿ tremendously, in much the same way that the long multiplication algorithm speeds up multiplication over the slower method of repeated addition. The benefit is had for n greater than or equal to 4. (Note: when measured in terms of the size of the problem, this algorithm is linear time since the size of an integer is its logarithm.)

[edit] Further applications

The same idea allows fast computation of large exponents modulo a number. Especially in cryptography, it is useful to compute powers in a ring of integers modulo q. It can also be used to compute integer powers in a group, using the rule

Power(x, -n) = (Power(x, n))^-1.

The method works in every semigroup and is often used to compute powers of matrices,

For example, the evaluation of

13789⁷²²³⁴¹ (mod 2345)

would take a very long time and lots of storage space if the naïve method is used: compute 13789⁷²²³⁴¹ then take the remainder when divided by 2345. Even using a more effective method will take a long time: square 13789, take the remainder when divided by 2345, multiply the result by 13789, and so on. This will take 722340 modular multiplications. The square-and-multiply algorithm is based on the observation that 13789⁷²²³⁴¹ = 13789(13789²)³⁶¹¹⁷⁰. So, if we computed 13789², then the full computation would only take 361170 modular multiplications. This is a gain of a factor of two. But since the new problem is of the same type, we can apply the same observation again, once more approximately halving the size.

The repeated application of this algorithm is equivalent to decomposing the exponent (by a base conversion to binary) into a sequence of squares and products: for example

x¹³ = x^1101_bin

= x^{(1*2^3 + 1*2^2 + 0*2^1 + 1*2^0)}

= x^{1*2^3} * x^{1*2^2} * x^{0*2^1} * x^{1*2^0}

= x^{2^3} * x^{2^2} * 1 * x^{2^0}

= x⁸ * x⁴ * x¹

= (x⁴)² * (x²)² * x

= (x⁴ * x²)² * x

= ((x²)² * x²)² * x

= ((x² * x)²)² * x       → algorithm needs only 5 multiplications instead of 13 - 1 = 12

Some more examples:

• x¹⁰ = ((x²)²*x)² because 10 = (1,010)₂ = 2³+2¹, algorithm needs 4 multiplications instead of 9
• x¹⁰⁰ = (((((x²*x)²)²)²*x)²)² because 100 = (1,100,100)₂ = 2⁶+2⁵+2², algorithm needs 8 multiplications instead of 99
• x^1,000 = ((((((((x²*x)²*x)²*x)²*x)²)²*x)²)²)² because 10³ = (1,111,101,000)₂, algorithm needs 14 multiplications instead of 999
• x^1,000,000 = ((((((((((((((((((x²*x)²*x)²*x)²)²*x)²)²)²)²)²*x)²)²)²*x)²)²)²)²)²)² because 10⁶ = (11,110,100,001,001,000,000)₂, algorithm needs 25 multiplications
• x^{1,000,000,000} = ((((((((((((((((((((((((((((x²*x)²*x)²)²*x)²*x)²*x)²)²)²*x)²*x)²)²*x)²)²*x)²*x)²)²)²*x)²)²*x)²)²)²)²)²)²)²)²)² because 10⁹ = (111,011,100,110,101,100,101,000,000,000)₂, algorithm needs 41 multiplications

[edit] Example implementations

[edit] Computation by powers of 2

This is a non-recursive implementation of the above algorithm in the Ruby programming language.

In most statically typed languages, result=1 must be replaced with code assigning an identity matrix of the same size as x to result to get a matrix exponentiating algorithm. In Ruby, thanks to coercion, result is automatically upgraded to the appropriate type, so this function works with matrices as well as with integers and floats.

def power(x,n)
  result = 1
  while n.nonzero?
    if n.modulo(2).nonzero?
      result = result * x
      n = n-1
    else
      x = x*x
      n = n/2
    end
  end
  return result
end

[edit] Runtime example: Compute 3¹⁰

parameter x =  3
parameter n = 10
result := 1

Iteration 1
  n = 10 -> n is even
  x := x2 = 32 = 9
  n := n / 2 = 5

Iteration 2
  n = 5 -> n is odd
      -> result := result * x = 1 * x = 1 * 32 = 9
         n := n - 1 = 4
  x := x2 = 92 = 34 = 81
  n := n / 2 = 2

Iteration 3
  n = 2 -> n is even
  x := x2 = 812 = 38 = 6561
  n := n / 2 = 1

Iteration 4
  n = 1 -> n is odd
      -> result := result * x = 32 * 38 = 310 = 9 * 6561 = 59049
         n := n - 1 = 0

return result

[edit] Runtime example: Compute 3¹⁰ =

result := 3
bin := "1010"

Iteration for digit 2:
  result := result2 = 32 = 9
  1010bin - Digit equals "0"

Iteration for digit 3:
  result := result2 = (32)2 = 34  = 81
  1010bin - Digit equals "1" --> result := result*3 = (32)2*3 = 35  = 243

Iteration for digit 4:
  result := result2 = ((32)2*3)2 = 310  = 59049
  1010bin - Digit equals "0"

return result

JavaScript-Demonstration: http://home.arcor.de/wzwz.de/wiki/ebs/en.htm

[edit] Generalization with example

[edit] Generalization

Let the pair ( S, * ) be a Semigroup, that means S is an arbitrary set and * is an associative binary operation on S:

• For all elements a and b of S is a*b also an element of S
• For all elements a, b and c of S is valid: (a*b)*c equals a*(b*c)

We may call * a "multiplication" and define an "exponentiation" E in the following way:
For all elements a of S:

• E ( a, 1 ) := a
• For all natural numbers n > 0 is defined: E ( a, n+1 ) := E ( a, n ) * a

Now the algorithm exponentiation by squaring may be used for fast computing of E-values.

[edit] Text application

Because the concatenation + is an associative operation on the set of all finite strings over a fixed alphabet ( with the empty string "" as its identity element ) exponentiation by squaring may be used for fast repeating of strings.

Example ( javascript ):
function repeat ( s, n ) {
  if ( s == "" || n < 1 ) return ""
  var res = s
  var bin = n.toString ( 2 )
  for ( var i = 1 ; i < bin.length ; i++ ) {
    res = res + res
    if ( bin.charAt ( i ) == '1' ) res = res + s
  }
  return res
}
The call repeat ( 'Abc', 6 ) returns the string AbcAbcAbcAbcAbcAbc

[edit] Calculation of products of powers

Exponentiation by squaring may also be used to calculate the product of 2 or more powers. If the underlying group or semigroup is commutative then it is often possible to reduce the number of multiplication by computing the product simultaneously.

[edit] Example

The formula a⁷×b⁵ may by calculated within 3 steps:

((a)²×a)²×a (four multiplications for calculating a⁷)

((b)²)²×b (three multiplications for calculating b⁵)

(a⁷)×(b⁵) (one multiplication to calculate the product of the two)

so one gets eight multiplications in total.

A faster solution is to calculate both powers simultaneously:

((a×b)²×a)²×a×b

which needs only 6 multiplications in total. Note that a×b is calculated twice, the result could be stored after the first calculation which reduces the count of multiplication to 5.

Example with numbers:

2⁷×3⁵ = ((2×3)²×2)²×2×3 = (6²×2)²×6 = 72²×6 = 31,104

Calculating the powers simultaneously instead of calculating them separately always reduces the count of multiplications iff at least two of the exponents are greater than 1.

[edit] Using transformation

The example above a⁷×b⁵ may also be calculated with only 5 multiplications if the expression is transformed before calculation:

a⁷×b⁵ = a²×ab⁵ with ab := a×b

ab := a×b (one multiplication)

a²×ab⁵ = ((ab)²×a)²×ab (four multiplications)

Generalization of transformation shows the following scheme:
For calculating a^A×b^B×...×m^M×n^N
1st: define ab := a×b, abc = ab×c, ...
2nd: calculate the transformed expression a^A-B×ab^B-C×...×abc..m^M-N×abc..mn^N

Transformation before calculation often reduces the count of multiplications but in some cases it also increases the count (see the last one of the examples below), so it may be a good idea to check the count of multiplications before using the transformed expression for calculation.

[edit] Examples

For the following expressions the count of multiplications is shown for calculating each power separately, calculating them simultaneously without transformation and calculating them simultaneously after transformation.

Example: a⁷×b⁵×c³
separate: [((a)²×a)²×a] × [((b)²)²×b] × [(c)²×c] ( 11 multiplications )
simultaneous: ((a×b)²×a×c)²×a×b×c ( 8 multiplications )
transformation: a := 2 ab := a×b abc := ab×c ( 2 multiplications )
calculation after that: (a×ab×abc)²×abc ( 4 multiplications ⇒ 6 in total )

Example: a⁵×b⁵×c³
separate: [((a)²)²×a] × [((b)²)²×b] × [(c)²×c] ( 10 multiplications )
simultaneous: ((a×b)²×c)²×a×b×c ( 7 multiplications )
transformation: a := 2 ab := a×b abc := ab×c ( 2 multiplications )
calculation after that: (ab×abc)²×abc ( 3 multiplications ⇒ 5 in total )

Example: a⁷×b⁴×c¹
separate: [((a)²×a)²×a] × [((b)²)²] × [c] ( 8 multiplications )
simultaneous: ((a×b)²×a)²×a×c ( 6 multiplications )
transformation: a := 2 ab := a×b abc := ab×c ( 2 multiplications )
calculation after that: (a×ab)²×a×ab×abc ( 5 multiplications ⇒ 7 in total )

[edit] Implementation

// the following javascript function calculates
// Bas [0] ^ Exp [0] x Bas [1] ^ Exp [1] x ...
function productOfPowers_simpleVersion ( Bas , Exp ) {
  var str // temporary string
// make binary representations:
  var maxLen = 0
  var bin = new Array ()
  for ( var i = 0 ; i < Exp.length ; i++ ) {
    str = Exp [i] . toString ( 2 )
    bin [i] = str
    if ( maxLen < str.length ) maxLen = str.length
  }
// make all binaries the same length:
  for ( var i = 0 ; i < bin.length ; i++ ) {
    while ( bin [i] . length < maxLen ) bin [i] = '0' + bin [i]
  }
// calculate:
  var result = 1
// . use first binary digits:
  for ( var y = 0 ; y < bin.length ; y++ ) {
    str = bin [y]
    if ( str.charAt ( 0 ) == '1' ) {
      if ( result == 1 ) result = Bas [y] ; else result = result * Bas [y]
    }
  }
// . use remaining digits:
  for ( var x = 1 ; x < maxLen ; x++ ) { // x : all digits except first one
    result = result * result
    for ( var y = 0 ; y < bin.length ; y++ ) { // y : all factors
      str = bin [y]
      if ( str.charAt ( x ) == '1' ) result = result * Bas [y]
    }
  }
// ready:
  return result
}
//
// for the following function input has to be sorted:
// Exp [0] >= Exp [1] >= ...
function productOfPowers_withTransformation ( Bas , Exp ) {
// new bases:
  var tempBas = new Array ()
  tempBas [0] = Bas [0]
  for ( var i = 1 ; i < Bas.length ; i++ ) tempBas [i] = Bas [i] * tempBas [i-1]
// new exponents:
  var tempExp = new Array ()
  for ( var i = 0 ; i < Exp.length - 1 ; i++ ) tempExp [i] = Exp [i] - Exp [i+1]
  tempExp [Exp.length-1] = Exp [Exp.length-1]
// now compress:
  var basTrans = new Array ()
  var expTrans = new Array ()
  for ( var i = 0 ; i < tempExp.length ; i++ ) if ( tempExp [i] > 0 ) {
    basTrans.push ( tempBas [i] )
    expTrans.push ( tempExp [i] )
  }
// ready:
  return productOfPowers_simpleVersion ( basTrans , expTrans )
}
// now let's test it:
alert ( 'S1: ' + productOfPowers_simpleVersion      ( [ 2 , 3 ] , [ 7 , 5 ] ) ) // should be 31,104
alert ( 'T1: ' + productOfPowers_withTransformation ( [ 2 , 3 ] , [ 7 , 5 ] ) ) // once again: 31,104
alert ( 'T2: ' + productOfPowers_withTransformation ( [ 2 , 5 , 3 ] , [ 4 , 3 , 2 ] ) ) // 18,000
alert ( 'T3: ' + productOfPowers_withTransformation ( [ 2 , 5 , 3 ] , [ 3 , 3 , 2 ] ) ) //  9,000
alert ( 'T4: ' + productOfPowers_withTransformation ( [ 2 , 5 , 3 ] , [ 4 , 3 , 3 ] ) ) // 54,000
alert ( 'T5: ' + productOfPowers_withTransformation ( [ 2 , 5 , 3 ] , [ 3 , 3 , 3 ] ) ) // 27,000

[edit] Alternatives

Addition chain exponentiation can in some cases require fewer multiplications by using an efficient addition chain to provide the multiplication order. However, exponentiating by squaring is simpler to set up and typically requires less memory.